Question

Find the values of λ for which the determinant is zero.

Answer 1

Answer:

Determinant are special number that can only be defined for square matrices.

Step-by-step explanation:

Determinant are particularly important for analysis. The inverse of a matrix exist, if the determinant is not equal to zero.

How to find determinant

For a 2×2 matrix

$det ( \left[\begin{array}{cc}x&y\\a&z\end{array}\right] ) = xz-ay$

For a 3×3 matrix

we first decompose it to 2×2

$det (\left[\begin{array}{ccc}k&l&m\\o&p&q\\r&s&t\end{array}\right] )\\\\= k*det(\left[\begin{array}{cc}p&q\\s&t\end{array}\right] ) - l*det(\left[\begin{array}{cc}o&q\\r&t\end{array}\right] ) + m*det(\left[\begin{array}{cc}o&p\\r&s\end{array}\right] ) \\\\=k(pt-sq) - l(ot-rq) + m(os-rp)$

Example

Find the values of λ for which the determinant is zero

$\left[\begin{array}{ccc}s&-1&0\\-1&s&-1\\0&-1&1\end{array}\right]$

$det(\left[\begin{array}{ccc}s&-1&0\\-1&s&-1\\0&-1&1\end{array}\right])\\\\= s*det(\left[\begin{array}{cc}s&-1\\-1&1\end{array}\right] ) - (-1)*det(\left[\begin{array}{cc}-1&-1\\0&1\end{array}\right] ) + 0*det(\left[\begin{array}{cc}-1&s\\0&-1\end{array}\right] )\\\\= s(s(1)-(-1*-1)) - (-1)(-1*1 - (-1*0)) + 0\\= s(s - 1)) + 1(-1 + 0) \\=s^{2} -s-1$

Equating the determinant to zero

$s^{2} -s-1 =0$

s = $\frac{1}{2}$ * (1 ±5 )

s = 1.61 or -0.61